144
ANALYTICAL MECHANICS
3. Find the center of mass of a semicircular lamina. Selecting the coordinates and the element of mass as shown in Fig.' we have
dm = cr-pd6 ğdp,
nay crp dp d6 _ >_____________
V~ C'C'rpdpdB
Ğ/0 J 0
O
FIG. 76.
= 0.
PROBLEMS.
Find the center of mass of the lamina bounded by the following curves:
(1) y = mx, y mx, and y a.
(2) y = asi&x, y = 0, x = 0, and x = TT.
(3) yz = ax and #2 = by.
(4) x2 + 7/2 = a2, Z = 0, and y = 0.
(5) bV + a22/2 = a262, x = 0, and a/ = 0.
(6) r=a(l+cos0).
(7) r = a, 0 = 0, and 6 = 0.
(8) r = a, r = b, 6 = 0, anc
7T
V
116. Center of Mass of a Homogeneous Solid of Revolution. Let Fig. 77 represent any solid . obtained by revolving a plane curve about the x-axis. Then the center of mass lies on the axis of revolution. The position of the center of mass is found most conveniently when the element of mass is a thin slice obtained by two transverse sections.. The expression for the mass of such an element is
dm = r wy2 - dx,
where r is the density of the solid, y the radius of the slice,, and dx its thickness.